5V regulator with 7805

[ElectroNewby]

I have a few simple questions regarding this design we've had in school.
Sorry, haven't touched electronics in a while.
Mostly, it's the caps values I'm interested with. I'd like to know how they came up with the values.

1)
C1
It's an electrolytic 10uF rated 10V.
I know it's to supply a clean and steady voltage to the 7805 but why 10 microfarads ?
And why 10V if the source is 12V ? I would put a 16V there, am I wrong or the voltage drops of the diode/LED/7805 take care of that?

2)
C2
A ceramic/disc 0.1uF (it's written 104 on it). It's some sort of filter I beleive, right...
Why 0.1uF ?
Why not an electrolytic cap ?
What's it's use if I already have a cap to have a clean input (C1) ?

3) with the 1k resistor
(Not couting the voltage drop, but I should) can I "V=RI" the source and output ?
I:12v/1k=12mA and 12V*12mA= 144mW so O:144mW/5V=29mA ?


4) how would you improve it ?

Thanks !
**broken link removed**

 

[AnalogKid]

1. Sometimes there is a minimum value on the datasheet. The cap's job is to lower the input impedance that the regulator sees, and in particular newtralize the inductance of the wires going to the power source. This inductance can cause the regulator to oscillate. The 10 V rating is not just wrong, but stupid. Good design practice is for a capacitor to be rated at twice the voltage it will see, to I'd start with a minimum of 100 uF at 25 V in parallel with a 0.1 uF..

2. Some regulators require a minimum output capacitance, and some actually have a maximum capacitance that should not be exceeded. So it is a bit more complex than just piling on capacitance. In general, the output capacitor aids in the regulator's response to a load transient, a rapid change in the current being drawn from the regulator. It also filters the noise produced by the regulator circuit. Again, 100 uF in parallel with a 0.2 uF is usually a safe starting point.

3. The LED is installed backwards, another reason to question the quality of this schematic and wherever it came from.

3. The LED current and power does not affect the regulator circuit in any way. The regulator circuit draws some power from the source even when it has no load, but that power is definded on the datasheet and is not affected by other stuff at the input.

Also, why is there a 5 V battery connected to the regulator output? If this is supposed to indicate that the regulator output functions as a 5 V battery, it is a horrible way to show it. An electronic circuit schematic is a precise language, and this is very bad grammar.

ak

 

[KeepItSimpleStupid]

1) A "rule of thumb" is 1000 uF/Amp. 2x the voltage is good practice, so 10V is too small. There are other ways of figuring the capacitance value using ripp;e current.

2) C2 is known as a bypass cap. It has to be located close to the IC. Usually it's a value recommended by the manufacturer. Geberally, it's a reserve where the regulator can draw from and also to compensate for the inductance of the PCB trace. These bypass caps will be found on nearly all IC's. It's an art and sometimes multiple types are used in parallel. Usually ceramic is used.

3) R should be something like R<= (12-2.1)/10 mA. The 2.1 id Vf of a LED and varies by color and has a range. 10 mA is a typical operating current for a LED.

D1 is reverse polarity protection at the input. You can add a diode between Vout and Vin as the datasheet suggests (also the LM317)

AK: missed some of your observations.

 

[AnalogKid]

AK: missed some of your observations.

???

 

[Mikebits]

2x the voltage is good practice

I was always told 3x the voltage?

 

[KeepItSimpleStupid]

AK:
Backwards LED; The battery on the 5V side.

Mikebits:
I guess it depends on how long you want the product to last. All I can tell you is that a 50V cap of a 50V supply doesn;t work. 75 with 85 V surge does, but that's with a ZNR clamp.

 

[Mikebits]

This is from Sparkfun.

De-Rating Rule of Thumb
While people may argue about what is acceptable for de-rating capacitors, I would suggest using capacitors whose voltage rating is at least 2-3 times that of the expected voltage. For example, if you plan on decoupling a 5V power line, use a capacitor with a rated voltage at 10V or above. I recommend above 3x for tantalum (or better yet, don’t use tantalum capacitors at all). Generally, it’s a bad day if you see this:

 

[MrAl]

Hi,

A cap voltage rating of 2 times the nominal voltage should be good enough really. You'll note that many manufucturers these days dont even meet that standard, for example a 16v cap for a 12v line.

The value of the cap is very application specific however. Note that with a 10uf cap a 1 amp current pulse out of the cap for 1us takes the voltage down by 0.1 volt, while a 1 amp current pulse for 10us takes the voltage down by 1.0 volts, given some distance from the power supply to the cap. So the value is dependent on the distance from the power supply as well as what the load is doing. 10uf is just a somewhat minimum value to get the design process started. Lab testing may show that more capacitance is needed, or not needed, given the particular application.

Oh yeah, the LED is drawn backwards ha ha